Problem: $f(x, y, z) = (y^2 + xz, zx^3, 2x + y)$ What is $\dfrac{\partial f}{\partial y}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $(2y, 0, 1)$ (Choice B) B $(x, x^3, 0)$ (Choice C) C $(2y + x + z, x^3 + 3zx^2, 3)$ (Choice D) D $(z, 3zx^2, 2)$
Solution: The partial derivative of a vector valued function is component-wise partial differentiation. $\begin{aligned} &f(x, y, z) = (f_0(x, y, z), f_1(x, y, z), f_2(x, y, z)) \\ \\ &f_x = \left( \dfrac{\partial f_0}{\partial x}, \dfrac{\partial f_1}{\partial x}, \dfrac{\partial f_2}{\partial x} \right) \\ \\ &f_y = \left( \dfrac{\partial f_0}{\partial y}, \dfrac{\partial f_1}{\partial y}, \dfrac{\partial f_2}{\partial y} \right) \\ \\ &f_z = \left( \dfrac{\partial f_0}{\partial z}, \dfrac{\partial f_1}{\partial z}, \dfrac{\partial f_2}{\partial z} \right) \end{aligned}$ Because we're taking a partial derivative with respect to $y$, we'll treat $x$ and $z$ as if they were constants. Therefore, $f_y = (2y, 0, 1)$.